Analysis-2 lecture schemes

Transcription

1 Anlysis-2 lecture schemes (with Homeworks) Csörgő István November, 204 A jegyzet z ELTE Informtiki Kr 204. évi Jegyzetpályáztánk támogtásávl készült

2 Contents. Lesson 4.. Continuity of functions Discontinuities Compct sets Homeworks Lesson Continuous functions defined on compct sets Continuous functions defined on intervls Monotone nd continuous functions defined on intervls The rel exponentil nd logrithm functions Homeworks Lesson Differentition of functions Some bsic derivtives Differentition Rules Some other bsic derivtives Homeworks Lesson Locl extrem of functions Men Vlue Theorems Discussion of Monotonity Inverse trigonometric functions Homeworks Lesson The L Hospitl Rule Tylor-polynomils Concvity Homeworks Lesson The Antiderivtive Five Simple Integrtion Rules Integrtion of Rtionl Functions Homeworks

3 CONTENTS 3 7. Lesson Integrtion by Prts Substitution Homeworks Lesson The definite Integrl Oscilltion Sum Bckwrd integrtion The properties of the definite integrl Homeworks Lesson Integrbility of continuous functions Piecewise continuous functions Integrl Function Lesson The Fundementl Theorem of Clculus (Newton-Leibniz) Integrtion by Prts Substitution Homeworks Lesson 55.. Improper Integrl Homeworks Lesson Applictions of the definite integrl Homeworks

4 . Lesson.. Continuity of functions Review: The neighbourhood (or bll) of the point x 0 R with rdius r > 0 is the set B(x 0, r) := {x R x x 0 < r } = (x 0 r, x 0 + r). Using this concept we cn define the continuity... Definition Let f R R, D f. f is continuous t df ε > 0 δ > 0 x B(, δ) D f : f(x) B(f(), ε). Let us denote the set of functions tht re continuous t by C(). From the definition it follows immeditely tht if is n isolted point of D f then f is continuous t. if is n ccumultion point of D f then f is continuous t lim x f(x) = f()..2. Definition Let f R R. The function f is clled to be continuous if it is continuous t every point of its domin, tht is D f : f C(). Using this observtion nd our knowledge bout the limit of functions, we cn stte tht the following functions re continuous t every point of their domin, so they re continuous functions: Constnt function, identity function, polynomils, rtionl functions (e. g.: /x), nlyticl functions (e. g.: exp, sin, cos, sh, ch)..3. Theorem [the Trnsference Theorem for continuity] Using our nottions: f C() x n D f (n N), lim x n = : lim f(x n ) = f(). The proof of the Trnsference Theorem is similr to tht of the cse of limit. Using the Trnsference Theorem it is esy to see tht f, g C(), c R f + g, f g, f g, f/g, c f C(), moreover g C(), f C(g()) f g C().

5 .2. Discontinuities 5.2. Discontinuities.4. Definition Let f R R D f. We sy tht f hs discontinuity t if f / C(). Remrk tht continuity nd discontinuity re defined only t the points of the domin nd re not defined t points outside of the domin. The discontinuities of n R R function re clssified in the following wy:.5. Definition Let f R R, D f, f / C(). We sy tht is point of removble discontinuity lim f, but lim f f(). jump lim f nd lim + f, but lim f lim + f. discontinuity of second kind lim f or lim + f..6. Exmples. The function f(x) = is continuous function. x if x 0 2. The function f(x) = x hs jump t 0. 0 if x = 0 3. The function f(x) = x 2 if x 0 0 if x = 0.3. Compct sets hs removble discontinuity t 0. Once more recll tht the neighbourhood (or bll) of the point x 0 R with rdius r > 0 is the set B(x 0, r) := {x R x x 0 < r } = (x 0 r, x 0 + r). Vi the concept of bll we cn define importnt clsses of points in connection of set..7. Definition Let = H R, x 0 R. Then. x 0 is n interior point of H, if r > 0 : B(x 0, r) H.

6 6. Lesson 2. x 0 is n exterior point of H, if r > 0 : B(x 0, r) H = tht is B(x 0, r) H. Here H denotes the complement of H tht is H = R \ H. 3. x 0 is boundry point of H, if r > 0 : B(x 0, r) H nd B(x 0, r) H..8. Remrk. Every interior point lies in H, every exterior point lies in H. But boundry point cn belong to H or to its complement..9. Definition. The set of the interior points of H is clled the interior of H nd is denoted by inth. So inth := {x 0 R r > 0 : B(x 0, r) H} H. 2. The set of the exterior points of H is clled the exterior of H nd is denoted by exth. So exth := {x 0 R r > 0 : B(x 0, r) H} H. 3. The set of the boundry points of H is clled the bound of H nd is denoted by H. So H := {x 0 R r > 0 : B(x 0, r) H nd B(x 0, r) H } R..0. Remrk. R = inth H exth (union of disjoint sets)... Definition Let H R. Then. H is clled n open set df H H. 2. H is clled closed set df H H..2. Remrks. So H is open if nd only if it does not contin ny boundry point nd is closed if nd only if it contins ll of its boundry points nd R re open nd closed sets t the sme time. There is no other set in R tht is open nd closed t the sme time. H is open H is closed, H is closed H is open. H is open H inth H = inth. About the chrcteriztion of closed sets we present the following theorem without proof:.3. Theorem Let = H R. Then H is closed if nd only if x n H (n N) convergent sequence : lim x n H. n

7 .3. Compct sets 7 After these preliminries we cn define the concept of compct sets..4. Definition Let = H R. H is clled compct set if x n H (n N) sequence (x νn ) subsequence : (x νn ) is convergent nd lim n x ν n H. The is clled to be compct by definition. Remrk tht from the definition it follows immeditely tht compct set is closed..5. Theorem Let = H R. Then H is compct if nd only if it is closed nd bounded. Proof. First suppose tht H is compct. Then H is closed s noted bove. Suppose indirectly tht H is unbounded. Then n N x n H : x n > n. By this wy we hve defined sequence x n H (n N). Tking subsequence (x νn ) we hve x νn > ν n n (n N). So (x νn ) is not bounded which implies tht it is not convergent. Therefore (x n ) does not contin convergent subsequence. Conversely, suppose tht H is closed nd bounded set nd let x n H (n N) be sequence in H. Then (x n ) is bounded so by the Bolzno-Weierstrss theorem (see: Anlysis-) it hs convergent subsequence (x νn ). Using tht H is closed we hve lim x ν n n H..6. Remrks. The theorem is vlid if we tke R n insted of R nd norm insted of bsolute vlue (see: Anlysis-3). In infinite dimensionl normed spces the theorem is not vlid. Every compct set is closed nd bounded but there exists closed nd bounded set in the spce tht is not compct (see: Functionl Anlysis). The following theorem is very importnt from the point of view of the extreme vlues of functions. Recll tht α = min H is miniml element of H if α H nd x H : x α. Respectively: β = mx H is mximl element of H if β H nd x H : x β..7. Theorem Let = H R be compct set in R. Then H hs miniml element min H nd mximl element mx H.

8 8. Lesson Proof. We will prove the cse mx H, the cse of minimum cn be proved similrly. H is compct H is bounded H is bounded bove α = sup H R. We need to prove tht α H. To show this use the fct tht for every n N the number α is not n n upper bound, so α is upper bound so we hve n N x n H : x n > α n. α n < x n α. Let n nd use the Sndwich Theorem (see: Anlysis-) to obtin: lim x n = α. n Since H is closed we hve α = lim x n H. n.4. Homeworks Discuss the continuity of the following functions (t which points of the domin is it continuous, t which points is it not, the type of the discontinuities, e.t.c x 2 x f(x) := 2 if x R\{2; 3} 5x if x {2; 3} (x 2) 2 f(x) := x 2 if x R\{2; 3} 5x if x {2; 3} x 2 if x 0 f(x) := ( x) 2 if 0 < x 2 4 x if x > 2 3 x if x 0, x 9 f(x) := 9 x 0 if x = 9

9 2. Lesson Continuous functions defined on compct sets 2.. Theorem [the continuous imge of compct set is compct] Let f R R be continuous function (f C) nd suppose tht D f compct. Then R f is compct. is Proof. Let y n R f (n N) be sequence in the rnge of f. Then x n D f : f(x n ) = y n (n N). D f is compct, so there exists convergent subsequence (x νn ) whose limit denoted by α is in D f. Using the Trnsference Theorem we obtin tht lim y ν n n = lim f(x ν n n ) = f(α) R f. So R f is compct. Before stting the following theorem let us define the extreme vlues of function: 2.2. Definition Let f R R. The minimum of f is the miniml element of its rnge (if exists) tht is min f := min R f = min {f(x) x D f } = min x D f f(x). Respectively, the mximum of f is the mximl element of its rnge (if exists) tht is mx f := mx R f = mx {f(x) x D f } = mx x D f f(x). These numbers re clled the bsolute (or globl) extreme vlues (bsolute (or globl) minimum, bsolute (or globl) mximum) of f Theorem [Theorem of Weierstrss] Let f R R, f C, D f compct. Then min f nd mx f. Proof. By the previous theorem R f is compct set in R, then by the.7 Theorem min R f nd mx R f. In the following definition we give stronger vrition of continuity, whose essence is tht the number δ in the definition of continuity is independent of the plce Definition Let f R R. We sy tht f is uniformly continuous if ε > 0 δ > 0 x, y D f, x y < δ : f(x) f(y) < ε.

10 0 2. Lesson Remrk. The definition of continuity of f mens tht y D f : f C(y) tht is y D f ε > 0 δ > 0 x D f, x y < δ : f(x) f(y) < ε. Compring these two definitions one cn see tht in the cse of continuity δ depends on both y nd ε tht is δ = δ(y, ε) but in the cse of uniform continuity δ depends only on ε nd is independent from the plce y tht is δ = δ(ε). Obviously every uniformly continuous function is continuous. The converse of this sttement is flse: there exists continuous but not uniformly continuous function. An exmple for such function is: f : (0, + ) R, x x. The following theorem tht is stted without proof gives sufficient condition for uniform continuity Theorem [Theorem of Heine bout the uniform continuity] Let f R R, f C, D f compct. Then f is uniformly continuous. The concept of uniform continuity on set is defined vi restriction: 2.7. Definition Let f R R, = H D f. We sy tht f is uniformly continuous on the set H if the restricted function f H is uniformly continuous Continuous functions defined on intervls Review: Let, b R, < b. The intervls with endpoints nd b re the wellknown sets: [, b] := {x R x b}: closed intervl; [, b) := {x R x < b}: intervl closed from the left nd open from the right; (, b] := {x R < x b}: intervl open from the left nd closed from the right; (, b) := {x R < x < b}: open intervl. Moreover we cn define the nonbounded intervls: [, + ) := {x R x }; (, + ) := {x R x > }; (, b] := {x R x b}; (, b) := {x R x < b}; (, + ) := R. The number is clled the left hnd endpoint (or: strting point) of the intervl nd the number b is clled the right hnd endpoint (or: terminl point) of the intervl. It cn be proved tht nonepty set H R is intervl if nd only if (inf H, sup H) H.

11 2.2. Continuous functions defined on intervls 2.8. Theorem [Intermedite Vlue Theorem, Theorem of Bolzno] Let f : [, b] R, f C. Suppose tht f() f(b), for exmple f() < f(b) (the discussion of the cse f() > f(b) is similr). Then c (f(), f(b)) ξ (, b) : f(ξ) = c. Proof. Let x 0 :=, y 0 := b, z := x 0 + y 0. 2 If f(z) = c, then ξ := z nd the proof is redy. If f(z) < c then x := z, y := y 0. If f(z) > c then x := x 0, y := z. For the intervl [x, y ] we hve [x, y ] [x 0, y 0 ], y x = y 0 x 0, f(x ) < c < f(y ). 2 Similrly we cn define recursively the intervl [x n+, y n+ ] from [x n, y n ]. So if the process does not stop t some step we obtin sequence of intervls ([x n, y n ]) for which x 0 x x 2... y 2 y y 0, y n x n = y 0 x 0 2 n nd f(x n ) < c < f(y n ) (n N). The sequence (x n ) is monotoniclly incresing nd bounded bove so it is converges to number α. Respectively the sequence (y n ) is monotoniclly decresing nd bounded below so it converges to the number β. Using the connection between the limit nd the ordering reltions follows: 0 β α y n x n = y 0 x 0 2 n 0 (n ). From here follows tht α = β =: ξ. The continuity of f t ξ implies using the Trnsference Theorem tht lim f(x n ) = lim f(y n ) = f(ξ). Using this fct let us mke n in the following inequlity f(x n ) < c < f(y n ) (n N) which implies f(ξ) c f(ξ) tht is f(ξ) = c Corollry. If f() f(b) < 0 then the eqution f(x) = 0 hs t lest one root in the intervl (, b) nd this root cn be pproximted by the sequences (x n ) nd (y n ) defined bove. The speed of the convergence is ( 2 )n (see: Numericl Anlysis) Theorem [The continuous imge of n intervl is intervl] Let f R R, f Cnd suppose tht D f is n intervl. Then R f is lso n intervl.

12 2 2. Lesson 2 Proof. It is enough to prove tht (inf R f, sup R f ) R f. Let inf R f < y < sup R f. By the definitions of sup nd inf it follows y R f : f(x ) = y < y nd y 2 R f : f(x 2 ) = y 2 > y, where x, x 2 re suitble elements in D f. f(x ) < y < f(x 2 ) therefore using Bolzno s theorem we obtin: ξ (x, x 2 ) I : f(ξ) = y. Consequently y R f Monotone nd continuous functions defined on intervls About the definition of monotonity nd bout the limit of monotone functions see: Anlysis Theorem Let I R be n intervl with strting point nd with terminl point b. Let f : I R be continuous. Suppose tht f is monotoniclly incresing ( ). Then the strting point of R f is lim f nd the terminl point of R f is lim f. + b Moreover if f is strictly monotoniclly incresing ( ) then its inverse function f is lso strictly monotoniclly incresing ( ) nd continuous Remrks.. If f is continuous nd strictly incresing ( ) then the following cses my occur: I = D f [, b] (, b] [, b) (, b) R f [f(), f(b)] (lim f, f(b)] + [f(), lim b f) (lim + f, lim b f) 2. A similr theorem cn be stted if f is monotoniclly decresing ( ) or strictly monotoniclly decresing ( ). The following theorem cn be proved lso by mens of Bolzno s theorem: 2.3. Theorem Let I R be n intervl, f : I R be continuous. Suppose tht f is one-to-one (injective). Then f is strictly monotone ( or ) function.

13 2.4. The rel exponentil nd logrithm functions The rel exponentil nd logrithm functions Review: In Anlysis- we hve lerned bout the exponentil function of rel or complex vrible. In this section let us look t the rel vrible cse: exp x := + x! + x2 2! + x3 3! +... = We hve seen in Anlysis- tht n=0 x n n! (x R). exp 0 =, exp(x + y) = exp x exp y, exp( x) = exp x. From the continuity of nlyticl functions it follows immeditely tht exp is continuous function. From the power series expnsion of exp the below properties follow immeditely:. x > 0 : exp x >. 2. x < 0 : 0 < exp x <. 3. lim exp x = +. x + 4. lim exp x = 0. x 2.4. Theorem exp : R R is strictly monotoniclly incresing ( ) function. Proof. Let x, y R, x < y. Then y x > 0 so exp(y x) >. Therefore exp y = exp ((y x) + x) = exp(y x) exp (x) > exp x = exp x. Using the 2. Theorem we hve tht R exp = (0, + ) Definition The inverse function of the rel exponentil function is clled nturl logrithm function nd is denoted by ln. So ln := exp. From this definition one cn simply deduce the following bsic properties of the nturl logrithmic function: D ln = R exp = (0, + ), R ln = D exp = R, ln = 0, ln (xy) = ln x + ln y, ln is strictly monotoniclly incresing, lim ln x = +, lim ln x =. x + x 0 0 In the next prt the connection between the rel exponentil function nd the powers of e is discussed.

14 4 2. Lesson Theorem r Q : exp r = e r ( where e denotes the Euler-number e = lim + n. n n) Proof. In Anlysis- we hve seen tht ( exp ( )) ( p q q = exp q i= = exp p = exp n=0 = e. We cn write for ny p, q N: n! ) ( p = exp q p q q ( p ) i= ) = = (exp ) p = e p. ( ) p So exp = e p/q, nd the theorem is proved for positive rtionl numbers. q In the cse r Q, r < 0 we write (using r > 0): exp r = exp( ( r)) = exp( r) = e r = er. The cse r = 0 is trivil Definition Let x R. Then the power e x is defined s e x := exp x Remrk. exp x is the unique continuous extension of e x from Q to R. This fct is bsed on the density of Q in R (see: Anlysis-) Definition Let > 0. Then the exponentil function with the bse is defined s follows: exp x := exp(x ln ) (x R). Remrk tht exp e x = exp x. It cn be proved similrly to the exp function tht r Q : exp r = r. This fct motivtes the definition x := exp x for every x R (unique continuous extension from Q to R). One cn see esily tht the function R x exp x is invertble if nd only if Definition Let > 0,. The inverse function of the exponentil function with the bse is clled the logrithm function with bse nd is denoted by log. So log := exp.

15 2.5. Homeworks 5 Remrk tht log e x = ln x. Finlly let us observe tht the power functions with fixed exponent µ R cn be written in the form: x µ = exp x µ = exp(µ ln x) (x R, x > 0). All the usul rules nd identities in connection with the powers nd logrithms tht we hve lerned in the secondry school cn be proved Homeworks Prove tht the given equtions hve roots in the given intervls. Is this root unique?. x 3 3x + = 0 in the intervl (0, ). Compute the first 3 terms of the sequence tht pproximtes the root. Estimte the error of pproximtion with this 3-rd term. 2. ln x = e x in the intervl (, e). 3. cos x = x in the intervl (0, ).

16 3. Lesson Differentition of functions 3.. Definition Let f R R, intd f. f is differentible t df f(x) f() lim R. x x In this cse f f(x) f() () := lim. This number is clled the derivtive of f x t the point. Let us denote the set of functions tht re differentible t by D() Remrks.. Other nottions for f () re: ( ) df, (f(x)) x= dx. x= 2. The geometricl mening of the derivtive is: the slope of the tngent line to the grph of f t the point (, f()). 3. The physicl mening of the derivtive is: the instntneous velocity of process (e.g. motion). Using the substitution h = x we obtin n equivlent form: f () := lim f(x) f() x f( + h) f() = lim. h 0 h The lst expression is useful becuse the letter x becme free so we cn write the definition so: f f(x + h) f(x) (x) = lim. h 0 h 3.3. Definition Let f R R nd suppose tht the set D f := {x intd f f D(x)} is nonempty. Then the function f : D f R, x f (x) is clled the derivtive function (or simply: the derivtive) of f. If D f = intd f then we sy tht the function f is differentible nd denote this fct by f D.

17 3.2. Some bsic derivtives Theorem f D() f C(). Proof. The difference f(x) f() tends to 0 since: f(x) f() = f(x) f() x (x ) f () 0 = 0 (x ) Remrk tht the opposite sttement is not true. For exmple let us tke the continuous function f(x) = x t 0: x So lim x 0 x x 0 lim x 0+ x 0 = lim x x 0+ x = lim x x 0+ x = lim =, x 0+ x 0 lim x 0 x 0 = lim x x 0 x = lim x x 0+ x = lim ( ) =. x 0+ does not exist, i.e. f / D(0) Some bsic derivtives In this section we compute some importnt bsic derivtives using the definition.. f(x) := c where c R is fixed (the constnt function). Then x R: f c c (x) = lim h 0 h = lim 0 = 0. h 0 2. f(x) = x + b where, b R re fixed (the liner function). Then x R: f (x + h) + b x b x + h + b x b (x) = lim = lim = lim =, h 0 h h 0 h h 0 especilly (x) =. 3. f(x) = x n where n N is fixed. Then using the binomil theorem x R: f (x + h) n x n (x) = lim = h 0 ( h n = lim 0) x n + ( ) n x n h + ( ) n 2 x n 2 h ( ) n n h n x n = h 0 (( ) ( ) h ( ) ) n n n = lim x n + x n 2 h h n = h 0 2 n ( ) n = x n = n x n.

18 8 3. Lesson 3 4. f(x) = e x (the exponentil function). Then x R: f (x) = lim h 0 e x+h e x h = lim h 0 e x e h e x h = e x e h lim = e x = e x. h 0 h = lim e x eh = h 0 h e x Here we hve used the fmilir limit lim = (see: Prctice). x 0 x 5. f(x) = sin x (the sinus function). Then x R: f sin(x + h) sin x sin x cos h + cos x sin h sin x (x) = lim = lim h 0 h h 0 ( h = lim cos x sin h h 0 h sin x cos h ) h 2 h = cos x. = sin x Here we hve used the fmilir limits lim x 0 x (see: Prctice). 6. (cos x) = sin x (x R) cn be proved similrly. cos x = nd lim x 0 x 2 = Differentition Rules. Sum 3.5. Theorem Let f, g D(x). Then f + g D(x) nd (f + g) (x) = f (x) + g (x). Proof. It cn be proved tht x intd f+g. To see the derivtive of f + g let us compute s follows: (f + g) (f + g)(x + h) (f + g)(x) (x) = lim = h 0 h f(x + h) + g(x + h) f(x) g(x) = lim = h 0 h f(x + h) f(x) g(x + h) g(x) = lim + lim = f (x) + g (x). h 0 h h 0 h 2. Product

19 3.3. Differentition Rules Theorem Let f, g D(x). Then fg D(x) nd (fg) (x) = f (x) g(x) + f(x) g (x). Proof. It cn be proved tht x intd fg. To see the derivtive of fg let us compute s follows: (fg) (fg)(x + h) (fg)(x) f(x + h) g(x + h) f(x) g(x) (x) = lim = lim = h 0 h h 0 h f(x + h) g(x + h) f(x) g(x + h) + f(x) g(x + h) f(x) g(x) = lim = h 0 h f(x + h) f(x) g(x + h) g(x) = lim lim g(x + h) + f(x) lim = h 0 h h 0 h 0 h = f (x)g(x) + f(x)g (x). The sclr multiple specil cse: Apply the Product Rule with the constnt function g(x) = c to obtin: (c f(x)) = c f (x). 3. Quotient 3.7. Theorem Let f, g D(x), g(x) 0. Then f g D(x) nd ( ) f (x) = f (x) g(x) f(x) g (x) g (g(x)) 2. Proof. It cn be proved tht x intd f/g. To see the derivtive of f g let us compute s follows: ( ) f ( f g (x) = lim )(x + h) ( f g )(x) f(x+h) = lim g h 0 h h 0 h f(x + h) g(x) f(x) g(x + h)) = lim = h 0 h g(x) g(x + h) g(x+h) f(x) g(x) f(x + h) g(x) f(x) g(x) + f(x) g(x) f(x) g(x + h) = lim = h 0 h g(x) g(x + h) [ ] [ ] f(x + h) f(x) g(x + h) g(x) = lim lim g(x) f(x) lim = h 0 g(x) g(x + h) h 0 h h 0 h = g(x) g(x) [f (x) g(x) f(x) g (x) ] = = f (x) g(x) f(x) g (x) (g(x)) 2. =

20 20 3. Lesson 3 Specil cse: the reciprocicl: Apply ( ) the Quotient Rule with the constnt function f(x) =. We obtin: = g (x) g(x) (g(x)) Composition (Chin Rule) without proof 3.8. Theorem Let g R R, g D(x), f R R, f D(g(x)). Then f g D(x) nd (f g) (x) = f (g(x)) g (x). 5. Inverse Rule without proof 3.9. Theorem Let I R be n open intervl, f : I R, f D, be n (strictly) incresing function. Furthermore suppose tht f (x) 0 (x I). Then f D(J) where J = R f (we know tht R f is n open intervl) nd (f ) (y) = f (f (y)) (y J) Some other bsic derivtives Using the Differentition Rules we cn deduce the derivtives of some bsic functions.. f(x) := tg x (the tngent function). Then using the Quotient Rule x D tg = R \ { π 2 + k π k Z}: tg x = ( ) sin x = sin x cos x sin x cos x cos x cos 2 = cos2 x + sin 2 x x cos 2 = x cos 2 x = +tg 2 x. 2. g(x) := ln x (x > 0) (the nturl logrithm function). Let f(x) = e x = exp x (x R). Then f (y) = ln y ssumptions of the Inverse Rule re stisfied, so: ln (y) = (f ) (y) = exp (ln (y)) = exp (ln (y)) = y (y R + ). All the (y J = R + ). If y is exchnged for x : (ln x) = x (x R+ ).

21 3.5. Homeworks Homeworks. Compute by definition the derivtive of f(x) = 2x t the point x 0 = Determine the derivtives of ) f(x) = 4x + 3 x b) f(x) = ln tg sin cos x c) f(x) = (x 2 + 2) sin x + 3 d) f(x) = tg x + tg 2 x 3. Determine the eqution of the tngent line to the given curve t its given point (only the first coordinte x 0 of the point is given): y = ln 2 (x, x 0 = 2. x )

22 4. Lesson Locl extrem of functions In connection with the Weierstrss-theorem (see: 2.3 Theorem) we hve defined the (globl or bsolute) extreme vlues of function. Now we will discuss the so clled locl extrem. 4.. Definition Let f R R, D f. We sy tht f hs t. locl minimum df r > 0 x B(, r) D f : f(x) f(); 2. strict locl minimum df r > 0 x B(, r) D f \ {} : f(x) > f(); 3. locl mximum df r > 0 x B(, r) D f : f(x) f(); 4. strict locl mximum df r > 0 x B(, r) D f \ {} : f(x) < f(); Here is the point of the locl extremum nd f() is the locl extreme vlue Theorem [First Derivtive Test for locl extremum] Let f R R, f D() nd suppose tht f hs locl extremum t. Then f () = 0. Proof. Suppose indirectly tht f () 0. Then either f () > 0 or f () < 0. Tke for exmple the cse f () > 0 (the other cse cn be discussed similrly). By the definition of the derivtive: 0 < f f(x) f() () = lim. x x It follows from the definition of the limit tht δ > 0 x ( δ, + δ) \ {} : f(x) f() x > f () 2 > 0. Since ( δ, + δ) \ {} = ( δ, ) (, + δ) let us discuss two cses: x <, x >. First let x ( δ, ). In this cse x < 0, so from the sign of frction follows tht f(x) f() < 0 tht is f(x) < f(). Similrly, if x (, + δ) then x > 0, so by the sign of the frction f(x) f() > 0 tht is f(x) > f(). Since ny neigbourhood of contin both types of these points the function f hs no extreme vlue t.

23 4.2. Men Vlue Theorems Remrks.. The reverse of the theorem is not true, see for exmple the function f(x) = x 3 (x R) t = If f R R, f D then the points of locl extrem re contined in the set of the roots of the eqution f (x) = 0. The roots of f (x) = 0 re clled criticl points or sttionry points Men Vlue Theorems 4.4. Theorem [Rolle] Let f : [, b] R, f C, f D. Suppose tht f() = f(b). Then ξ (, b) : f (ξ) = 0. Proof. By the Weierstrss-theorem min f nd mx f. If min f = mx f then f is constnt so every ξ (, b) is good choice. If min f < mx f then using f() = f(b) one of them is tken in the inside of [, b] tht is t certin ξ (, b). So the First Derivtive Test cn be pplied: f (ξ) = Theorem [Cuchy] Let f, g : [, b] R, f, g C, f, g D. Suppose tht g (x) 0 (x (, b)). Then ξ (, b) : f(b) f() g(b) g() = f (ξ) g (ξ). Proof. Let F (x) = f(x) + λg(x). We wnt to pply the Rolle-theorem for F therefore we choose the prmeter λ so ht F () = F (b) holds. F () = f() + λg() = f(b) + λg(b) = F (b) λ = f(b) f() g() g(b) = f(b) f() g(b) g(). By the Rolle-theorem ξ (, b) : F (ξ) = 0. So f (ξ) f(b) f() g(b) g() g (ξ) = 0. The sttement of the theorem cn be obtined by rerrnging this eqution Theorem [Lgrnge] Let f : [, b] R, f C, f D. Then ξ (, b) : f(b) f() b = f (ξ). Proof. Let us pply the Cuchy-theorem with g(x) = x (x [, b]): ξ (, b) : f(b) f() b = f(b) f() g(b) g() = f (ξ) g (ξ) = f (ξ) = f (ξ).

24 24 4. Lesson Discussion of Monotonity The monotonity of functions defined on intervls cn be effectively discussed using derivtives Theorem [First Derivtive Test for monotonity] Let I R be n intervl (of ny type), f : I R, f C, f D. Then. If x inti : f (x) > 0 then f is strictly incresing (on I). 2. If x inti : f (x) < 0 then f is strictly decresing (on I). 3. If x inti : f (x) = 0 then f is constnt (on I). Proof. Let x, x 2 I, x < x 2 nd let us pply the Lgrnge-theorem on the closed intervl [x, x 2 ]: ξ (x, x 2 ) : f(x 2 ) f(x ) x 2 x = f (ξ). After rerrngement: f(x 2 ) f(x ) = f (ξ) (x 2 x ).. Since ξ inti we hve f (ξ) > 0. Then x 2 x > 0 implies f(x 2 ) f(x ) > 0 tht is f(x ) < f(x 2 ). 2. Since ξ inti we hve f (ξ) < 0. Then x 2 x > 0 implies f(x 2 ) f(x ) < 0 tht is f(x ) > f(x 2 ). 3. Since ξ inti we hve f (ξ) = 0. Therefore f(x 2 ) f(x ) = 0 tht is f(x ) = f(x 2 ) Remrk. The prcticl ppliction of the theorem: Let f R R, D f be intervl. Suppose tht the eqution f (x) = 0 hs finite mny roots: x < x 2 <... < x k. If f is continuous then the sign of f is constnt on the intervl (x j, x j ). Consequently the function is strictly incresing or strictly decresing over [x j, x j ] Inverse trigonometric functions 4.9. Theorem There exists unique number α (0, 2) such tht cos α = 0.

25 4.4. Inverse trigonometric functions 25 Proof. For the existence it is enough to prove tht cos 0 > 0 nd cos 2 < 0. From here using the continuity of cos nd the Bolzno-theorem the existence follows. Indeed, cos 0 = > 0. On the other hnd: cos 2 = 22 2! ! 26 6! !... = 22 2! ! 26 6! 22n+2 22 ( )... }{{ 7 8} + (2n + 2)! ( 2 2 )... < 22 (2n + 3)(2n + 4) 2! ! = 3 < 0. }{{} + For the uniqueness it is enough to prove tht the cos function is strictly decresing over the intervl [0, 2]. Since the derivtive of cos is sin, it is enough to show tht sin x > 0 for ny 0 < x < 2. Relly, if 0 < x < 2 then sin x = x x3 3! + x5 5! x7 7! + x9 9! x! +... = = x ( x2 2 3 ) + x5 x2 ( 5! 6 7 ) + x9 x2 ( 9! 0 ) +... > > x ( 22 ) + x5 22 ( ) + x9 22 ( ) +... > 0. }{{ 2 3} 5! }{{ 6 7} 9! } 0 {{ } Definition Let π := 2α where α is the number in the previous theorem. Since α = π is the unique zero of cos in [0, 2] nd cos 0 = > 0 we cn hve 2 tht cos x > 0 if 0 x < π. The cos function is even, so cos x > 0 for every 2 π 2 < x < π 2. Since (sin x) = cos x there follows tht the sin function is strictly [ incresing over the intervl π 2, π ]. Let us compute sin π 2 2 : sin 2 x+cos 2 x = nd sin π 2 > 0 imply sin π 2 = cos 2 π 2 = 0 =. The sin function is odd, so sin( π 2 ) = sin π =. Therefore the rnge of the 2 restricted sin function is [, ]. [ π 2, π 2 ] By the previous fcts we cn stte tht the restricted sin [ π inverse which is clled rcsin. 2, π 2 ] function hs

26 26 4. Lesson 4 [ 4.. Definition rcsin := sin : [, ] π [ π 2, π 2 ] 2, π ] Remrk. rcsin y = x x [ π 2, π ] 2 nd sin x = y. The derivtive of rcsin: ( Let f(x) = sin x (x I := π 2, π ) ). In this cse f (y) = rcsin y (y 2 ( )). We cn pply the theorem bout the derivtive of inverse function: rcsin y = (f ) (y) = sin (rcsin y) = cos(rcsin y). ( Since rcsin y π 2, π ) it is obvious tht cos(rcsin y) > 0, so we cn continue 2 s follows: rcsin y = =. (sin(rcsin y)) 2 y 2 So rcsin y = y 2 (y (, )). After replcing y by x : rcsin x = x 2 (x (, )). Using similr considertion we cn define the inverses of cos, tg nd ctg. Here is the collection of them nd their derivtives: 4.3. Definition rccos := cos : [, ] [0, π] [0,π] ( rc tg := tg : R π ( π 2, π 2 ) 2, π ) 2 rc ctg := ctg : R (0, π) (0,π) Their derivtives cn be computed like the one of the rcsin. The results: rccos x = x 2 (x (, )) ; rc tg x = + x 2 (x R) ; rc ctg x = + x 2 (x R).

27 4.5. Homeworks Homeworks. On wht intervls is f incresing nd decresing? At wht points hs it locl extreme vlue? ) f(x) = x 3 3x 2 b) f(x) = x 2 (x ) 2 c) f(x) = x x 2 6x 6 d) f(x) = x e x 2. A right tringle whose hypotenuse is 3 long is revolved bout one of its legs to generte right circulr cone. Find the rdius nd height of the cone of gretest volume. 3. Find the bsolute extreme vlues (nd their plces) of f(x) = x x 2 + x + 4. Prove tht rc tg x = + x 2 (x R). ( 2 x 0).

28 5. Lesson The L Hospitl Rule An importnt ppliction of the differentil clculus is the computtion of indeterminte form limits vi the L Hospitl Rule: 5.. Theorem [L Hospitl Rule] Let < b +, f, g : (, b) R, f, g D, g (x) 0 (x (, b)). Suppose tht either lim f = lim g = 0 or lim f = lim g = + nd tht lim g. Then f lim +0 g = lim f +0 g. +0 f Proof. We will prove only tht prt of the cse 0 when >. 0 f Let A := lim +0 g nd ε > 0. This implies by the definition of limit: δ > 0 : + δ < b, x (, + δ) : f (x) g (x) B(A, ε). Let f() := g() := 0 nd tke number x (, + δ). Let us pply the Cuchy Men Vlue Theorem on the intervl [, x]: f(x) g(x) f(x) f() = g(x) g() = f (ξ) g B(A, ε). (ξ) It mens by the definition of limit tht the sttement of the theorem is true. The other cses of the theorem cn be proved similrly or cn be reduced to the proved cses. Similr theorem cn be proved for left-hnd limits. From here it follows tht similr theorem is vlid for limits. The indeterminte forms tht re not 0 0 or cn be reduced vi lgebric trnsforms to the indeterminte quotient cse Tylor-polynomils First we discuss the higher order derivtives. The second order derivtive is defined s the derivtive of the derivtive function.

29 5.2. Tylor-polynomils Definition Let f R R, intd f. We sy tht f is 2 times differentible t (its nottion is: f D 2 ()) if r > 0 x B(, r) : f D(x) nd f D(). In this cse the number f () := (f ) () is clled the second derivtive of f t the point. Similrly cn be defined the 3., 4.,... derivtives with recursion. Their nottions re: f (), f (),... or f (3) (), f (4) (),... Generlly if f is k times differentble t then we denote this fct by f D k () nd the k-th order derivtive by f (k) () Definition Let f R R nd suppose tht the set D f (k) := { x intd f } f D k (x) is nonempty. Then the function f (k) : D f (k) R, x f (k) (x) is clled the k-th order derivtive function (or simply: the k-th derivtive) of f. If D f (k) = intd f then we sy tht the function f is k times differentible nd denote this fct by f D k Definition (Tylor polynomil) Let f R R, f D n (). The polynomil T n (x) := f() + f () (x ) + f () (x ) f (n) () (x ) n =! 2! n! n f (k) () = (x ) k (x R) k! k=0 is clled the n-th Tylor-polynomil of f reltive to the center Remrks.. It is obvious tht the degree of T n is t most n tht is T n P n. 2. Obviously T n () = f(). 3. T n(x) = n k= f (k) () k! k (x ) k. Hence we hve T n() = f (). 4. Similrly using mthemticl induction one cn prove tht T (j) n () = f (j) () (j = 0,..., n).

30 30 5. Lesson Theorem [Tylor s formul] Let I R be n open intervl, f : I R, f D n+, I. Then for every x I \ {} there exists number ξ between nd x such tht: f(x) T n (x) = f (n+) (ξ) (n + )! (x ) n+. The right-hnd side of this eqution is clled the Lgrngin reminder term. Proof. Let us introduce the following uxiliry functions: F (z) := f(z) T n (z) nd G(z) := (z ) n+ (z I). One cn esily compute tht furthermore F (j) () = f (j) () T (j) n () = 0, G (j) () = 0 (j = 0,..., n), F (n+) (x) = f (n+) (x), G (n+) (x) = (n + )! (x I). Suppose tht x I, x > (right-hnd cse), nd let us pply the Cuchy Men Vlue Theorem consecutively (first for F nd G on the intervl [, x], then for F nd G on the intervl [, ξ ], etc.). Thus we obtin tht there exist numbers < ξ n+ < ξ n <... < ξ 2 < ξ < x such tht F (x) F (x) F () = G(x) G(x) G() = F (ξ ) G (ξ ) = F (ξ ) F () G (ξ ) G () = F (ξ 2 ) G (ξ 2 ) = =... = F (n) (ξ n ) G (n) (ξ n ) = F (n) (ξ n ) F (n) () G (n) (ξ n ) G (n) () = F (n+) (ξ n+ ) G (n+) (ξ n+ ) = f (n+) (ξ n+ ). (n + )! Let ξ := ξ n+. We hve obtined tht ξ (, x) : F (x) G(x) = f (n+) (ξ). (n + )! By the definitions of F nd G: f(x) T n (x) (x ) n+ = f (n+) (ξ). (n + )! From here we cn finish the proof by simple rerrngement. The left-hnd cse x < cn be proved similrly Remrk. In the cse n = 0, x = b > (where [, b] I) the Tylor s Formul coincides with the Lgrnge Men Vlue Theorem.

31 5.3. Concvity Concvity 5.8. Definition Let I R be n intervl, f : I R. () f is clled to be concve up (or: convex) if x, y I, x < y 0 < λ < : f(λx + ( λ)y) < λf(x) + ( λ)f(y) ; (b) f is clled to be concve down (or: concve) if x, y I, x < y 0 < λ < : f(λx + ( λ)y) > λf(x) + ( λ)f(y). The following theorem cn be proved: 5.9. Theorem Let I R be n intervl, f : I R, f C, f D. Then () f is concve up if nd only if f is strictly incresing; (b) f is concve down if nd only if f is strictly decresing. Remrk tht the sttement of the theorem could be the definition of concvity for such functions (f : I R, f C, f D). Using the First Derivtive Test for the monotonity of f we obtin 5.0. Theorem Let I R be n intervl, f : I R, f C, f D 2. Then () if x inti : f (x) > 0 then f is concve up; (b) if x inti : f (x) < 0 then f is concve down. On the grph of differentible function the points where the concvity chnges re of specil importnce. These points will be clled points of inflection. 5.. Definition Let I R be n intervl, f : I R, f C, f D 2, I. The point is clled point of inflection if exists number δ > 0 such tht ( δ, + δ I nd one of the following two cses holds: Cse : f ( δ,] is concve up nd f [,+δ) is concve down or Cse 2: f ( δ,] is concve down nd f [,+δ) is concve up Remrk tht the concept of point of inflection cn be defined in more generl form, but for our computtions the definition given bove will be pproprite Homeworks. Use the L Hospitl Rule to determine the following limits: ) lim x 0 sin x x rcsin x x b) lim x 0 x ctg (πx) ( c) lim x 0 x ) e x d) lim (cos x) x x 0+

32 32 5. Lesson 5 2. Mke complete discussion (with grphing) of the following functions: ) f(x) = 2 x 3 + x b) f(x) = x ln x 3. Let f(x) := + x (x ). ) Write the 2-nd degree Tylor s polynomil T 2 of f centered t 0. b) Estimte the error in the pproximtion f(x) T 2 (x).

33 6. Lesson The Antiderivtive In mny cses we look for function whose derivtive is given function. This process is clled ntidifferentition or indefinite integrtion. 6.. Definition Let I R be n open intervl, f : I R, F : I R. The function F is clled to be n ntiderivtive of f if F D nd x I : F (x) = f(x). About the set of ntiderivtives of function the following theorem holds: 6.2. Theorem Let I R be n open intervl, f : I R. Let F : I R be n ntiderivtive of f. Then the set of ll ntiderivtives of f is Proof. Since {I x F (x) + C C R} = {F + C C R}. (F (x) + C) = F (x) + C = f(x) + 0 = f(x) therefore the function F + C is indeed n ntiderivtive of f. On the other hnd let G : I R be n ntiderivtive of f. Since G = f then (G F ) = G F = f f = 0, so using the First Derivtive Test C R x I : (G F )(x) = C. After rerrngement: G(x) = F (x) + C, so G is of the form F + C Definition The set of ll ntiderivtives of the function f is clled the indefinite integrl of f nd is denoted by: f, f(x) dx. Remrk tht in the prctice sometimes the individul ntiderivtives re nmed lso indefinite integrl, for exmple: 3x 2 dx = x 3. The indefinite integrl in prctice is written not with set nottions but in the following wy: 3x 2 dx = x 3 + C. The next question tht we re concerned is: Which functions hve ntiderivtives? Lter we will prove the following theorem: 6.4. Theorem If I R is n open intervl nd f : I R is continuous function then f hs ntiderivtive.

34 34 6. Lesson Five Simple Integrtion Rules 6.5. Theorem [sum] Let I R be open intervl, f, g : I R. If f nd g hve ntiderivtives, so does f + g. Moreover f(x) + g(x) dx = f(x) dx + g(x) dx. Proof. ( f + g) = ( f) + ( g) = f + g Theorem [sclr multiple] Let I R be n open intervl, f : I R, λ R. If f hs ntiderivtive, so does λf, moreover λ f(x) dx = λ f(x) dx. Proof. (λ f) = λ ( f) = λ f Theorem [liner substitution] Let I R be n open intervl, f : I R nd F : I R be n ntiderivtive of f. Furthermore let, b R, 0 nd J := {x R x + b I}. Then J is n open intervl nd f(x + b) dx = F (x + b) Proof. Obviously J is n open intervl. Moreover: ( F (x + b) (x J). ) = F (x + b) = f(x + b). (x J) Theorem [integrls of type f α f ] Let I R be n open intervl, f : I R be continuous, α R. Suppose tht the power (f(x)) α is defined for every x I. Then ) if α then (f(x)) α f (x) dx = (f(x))α+ α + (x I);

35 6.3. Integrtion of Rtionl Functions 35 b) if α = then Proof. ) If α then: (f(x)) f (x) dx = f (x) f(x) dx = ln f(x) (x I). ( f α+ α + ) = α + (α + ) f α f = f α f ; b) If α = then: (ln f ) = f f = f f Integrtion of Rtionl Functions Using the previous simple integrtion rules we cn give method for integrtion of rtionl functions. Recll tht rtionl function is quotient of two polynomils. Let us study importnt bsic types first: Bsic type : Let A R, α R, k N +, I := (, α) or I := (α, + ) nd R(x) := A (x α) k (x I). In this cse using the rule of the liner substitution we obtin R(x) dx = A (x α) A k+ (x α) k dx = k + A ln x α,, if k 2; if k=. Bsic type 2: Let B, C R, β, γ R where β 2 4γ < 0, I := R nd R(x) := Bx + C x 2 + βx + γ (x I).

36 36 6. Lesson 6 In this cse we pply the following method. If B 0 then we introduce the derivtive of the denomintor into the numertor: Bx + C x 2 + βx + γ = B 2x + 2C 2 B x 2 + βx + γ = B 2x + β + 2C 2 B β x 2 = + βx + γ = B 2 2x + β x 2 + βx + γ + B 2 (2C B β) x 2 + βx + γ The first term is f type so its integrtion is esy. The second term is like the f originl function R but in the numertor the constnt stnds insted of the liner function Bx + C. If originlly B = 0 then fter seprting the fctor C we obtin the bove frction. So the problem is reduced to the form x 2 + βx + γ. The integrtion of this frction will be solved by eliminting the term βx (if β 0) in the denomintor by completing the squre. After this step we will trnsform the obtined frction into the form (x + b) 2 which using the + liner substitution nd the ntiderivtive of rc tg cn be integrted esily: x 2 + βx + γ = ( x + β ) 2 = β γ = 4 4γ β 2 ( γ β x β 2 γ β2 4 ) 2 2 x + β + 4γ β 2 4γ β 2 + = Bsic type 3: Let B, C R, β, γ R where β 2 4γ < 0, k N, k 2, I := R nd Bx + C R(x) := (x 2 + βx + γ) k (x I). In this cse using recursive process we trce the integrl of R bck to similr integrl but with exponent k insted of k. The recursive process is continued until the exponent k will be reduced to nd the problem becomes of Bsic type 2. The recursion process is bsed on the following theorem:

37 6.4. Homeworks Theorem There exist constnts B, C, D R such tht: Bx + C (x 2 + βx + γ) k dx = B x + C (x 2 + βx + γ) k + D (x 2 dx. + βx + γ) k Arbitrry rtionl functions: An rbitrry rtionl function cn be integrted by the method of prtil frction decomposition Theorem Let P nd Q be nonzero rel polynomils where the root fctor form of Q over R is: Q(x) = (x α ) m... (x α r ) mr (x 2 + β x + γ ) n... (x 2 + β s x + γ s ) ns. Here α,... α r re the rel roots of Q with multiplicities m,..., m r, β 2 i 4γ i < 0 (j =,... s), m m r + 2n n s = deg Q. Then R cn be written in form R(x) := P (x) Q(x) = S(x) + r m i i= j= A ij s (x α i ) j + n i i= j= where S is polynomil, A ij, B ij, C ij re rel coefficients. B ij x + C ij (x 2 + β i x + γ i ) j, So the integrl of rtionl frction is the sum of the integrl of polynomil nd of the integrls of some bsic type rtionl frctions Homeworks. Determine the ntiderivtives 5 ) cos 2 dx b) ( 6x + 4) c) e) 2x 2 5 (x 2)(x 2 ) dx d) x 2 (x )(x 2 + 2x + ) dx 2x 5 (x 2 5x + 3) dx 7 3 6x x 2 2x + 7 dx

38 7. Lesson Integrtion by Prts 7.. Theorem [integrtion by prts] Let I R be n open intervl, f, g : I R, f, g, D, f, g C. Then f(x) g (x) dx = f(x) g(x) f (x) g(x) dx (x I) Proof. Apply the product rule of derivtive: (f g (f g)) = (f g) ( (f g)) = f g + f g f g = f g Substitution 7.2. Theorem [Substitution, form I.] Let I, J R be open intervls, f : J R, f C, g : I J g D, g C. Then (f g) g = ( f) g. Proof. Apply the Chin Rule: (( f) g) = (( f) g) g = (f g) g Remrk. If the vrible of g is denoted by x then the rule of substitution cn be written s: f(g(x)) g (x) dx = F (g(x)) (x I), where F denotes n ntiderivtive of f. So denoting the vrible of f by u : F (u) = f(u) du (u J). From here cn we obtin the prcticl process of the substitution. In the integrl f(g(x)) g (x) dx substitute g(x) by u nd g (x) dx by du. After determintion of this new integrl substitute u by g(x).

39 7.3. Homeworks 39 Now suppose tht g is bijection (nturlly becuse of its continuity it is strictly monotone too). Then the bove rule of substitution cn be used in nother form: 7.4. Theorem [Substitution, form II.] Let I, J R be open intervls, f : I R, f C, g : J I, g D, g C. Then ( ) f = (f g) g g. Proof. In the substitution formul (form I.) let us interchnge the roles of I nd J: (f g) g = ( f) g. Then tke the composition of both sides with the function g : I J: ( ) (f g) g g = ( f) g g = f. After interchnge the sides of this equlity we obtin the desired formul Remrks.. If the vrible of f is denoted by x, the vrible of g is denoted by t then the form II. cn be written s: f(x) dx = f(g(t)) g (t) dt t=g (x) (x I). From here comes the prcticl process of substitution (form II.). In the integrl f(x) dx x is replced by g(t), dx is replced by g (t) dt. After computtion of this integrl replce t by g (x). 2. The function g cn be chosen freely. There re suggested substitutions for mny types of integrl, 3. If the conditions of the theorem bout the form II. hold then both forms of the substitution cn be used Homeworks. Determine the following integrls: x 2 ln x dx (x 2 + ) e 2x dx x cos x dx 2. Determine the following integrls: e x 9 4x dx 2 dx e 2x x x 3x + 5 dx

40 8. Lesson The definite Integrl 8.. Definition Let n N nd divide the intervl [, b] into n closed subintervls [x i, x i ] (i =,..., n) where: = x 0 < x < x 2 <... < x n = b. The finite set {x 0, x,..., x n } is clled prtition of the intervl [, b]. The set of the prtitions of [, b] will be denoted by P[, b] Definition Let P P[, b] Then the length of the longest subintervl is clled the norm of the prtition P : P := mx{x i x i i =,..., n}. It is obvious tht for every δ > 0 there exists prtition P finer thn δ tht is P < δ Definition Let f : [, b] R be bounded function nd P = {x 0, x,..., x n } P[, b]. Let m i := inf{f(x) x i x x i }, M i := sup{f(x) x i x x i } (i =,..., n). We introduce the following sums: ) lower sum: s(f, P ) := n m i (x i x i ), i= b) upper sum: S(f, P ) := n M i (x i x i ). i= 8.4. Theorem If P, Q P[, b], P Q then s(f, P ) s(f, Q) nd S(f, P ) S(f, Q) Corollry. If P, Q P[, b] then s(f, P ) s(f, P Q) S(f, P Q) S(f, Q) Corollry. The set of the lower sums is bounded bove, the set of the upper sums is bounded below Definition The number I (f) := sup{s(f, P ) P P[, b]} is clled the lower integrl of f. Respectively the number I (f) := inf{s(f, P ) P P[, b]} is clled the upper integrl of f.

41 8.. The definite Integrl Definition A function f : [, b] R is clled to be integrble if it is bounded nd I (f) = I (f). This common vlue of the lower nd upper integrl is clled the integrl of f from to b nd is denoted by f, f(x) dx. In this connection the number is clled the lower limit of the integrl nd the number b is clled the upper limit of the integrl. The definition cn be extended esily to the cse when the domin of f is wider thn [, b]: 8.9. Definition Let f R R, [, b] D f. We sy tht f is integrble over the intervl [, b] if the restricted function f [,b] is integrble. The integrl of f from to b is denoted by the previous wy nd is defined s f := f(x) dx := f [,b]. The set of integrble functions over [, b] is denoted by R[, b]. From the definition it follows tht in the cse f(x) 0 (x [, b]) the geometricl mening of the integrl is the re of the plnr region under the grph of f tht is the re of the region 8.0. Exmples R := { (x, y) R 2 x b, 0 y f(x) }.. Let c R be fixed nd f(x) := c (x [, b]) be the constnt function. Then for ny prtition P = {x 0, x,..., x n } of [, b]: n n s(f, P ) := c (x i x i ) = c (x i x i ) = c (b ), i= which implies tht I (f) = c (b ). On the other hnd n S(f, P ) := c (x i x i ) = c i= which implies tht I (f) = c (b ). So i= n (x i x i ) = c (b ), i= f(x) dx = I (f) = I (f) = c (b ).

42 42 8. Lesson 8 2. Here follows n exmple for nonintegrble (but bounded) function. Let f : [, b] R be the following function: if x Q [, b] f(x) := 0 if x (R \ Q) [, b]. Let P = {x 0, x,..., x n } be prtition of [, b]. Since every subintervl [x i, x i ] contins rtionl nd irrtionl numbers too we hve Thus m i := inf{f(x) x i x x i } = 0, M i := sup{f(x) x i x x i } = (i =,..., n). s(f, P ) := n m i (x i x i ) = 0, S(f, P ) := i= n M i (x i x i ) = b. i= Consequently I (f) = sup s(f, P ) = 0, I (f) = inf{s(f, P ) = b. P They re not equl, so f / R[, b] Oscilltion Sum 8.. Definition Let f : [, b] R be bounded function nd P = {x 0, x,..., x n } P[, b]. The number Ω(f, P ) := S(f, P ) s(f, P ) = n (M i m i ) (x i x i ) i= is clled oscilltion sum. (M i, m i were defined in the previous section.) The following theorem will be useful when we wnt to prove the integrbility of function Theorem Let f : [, b] R be bounded function. Then f R[, b] ε > 0 P P[, b] : Ω(f, P ) < ε.

43 8.3. Bckwrd integrtion 43 Proof. Assume tht f R[, b] nd let ε > 0. By the definition of the lest upper bound P P[, b] : s(f, P ) > I (f) ε 2. Similrly by the definition of the gretest lower bound So we cn write with P := P P 2 : P 2 P[, b] : S(f, P 2 ) < I (f) + ε 2. Ω(f, P ) := S(f, P ) s(f, P ) S(f, P 2 ) s(f, P ) < I (f)+ ε ( 2 I (f) ε ) = ε. 2 Conversely let ε > 0 be n rbitrry but fixed positive number nd P be prtition with Ω(f, P ) < ε. Then 0 I (f) I (f) S(f, P ) s(f, P ) = Ω(f, P ) < ε Since it is true for ny ε > 0 we infer tht I (f) is equl to I (f) Bckwrd integrtion It is convenient nd useful to extend the integrtion if its lower limit is greter thn or equl to its upper limit Definition Let f R R, f R[, b]. Then b f(x) dx := 8.4. Definition Let f R R, D f. Then f(x) dx := 0. f(x) dx. So we hve defined the definite integrl f(x) dx for ny pir, b R. For ny pir, b R let us denote the set of functions for which the integrl exists (independently of < b, = b, > b) by R[, b]. f(x) dx

44 44 8. Lesson The properties of the definite integrl In this section the theorems re stted without proofs Theorem [Addition] Let, b R, f, g R[, b]. Then f + g R[, b] nd (f + g) = f + g Theorem [Constnt Multiple] Let, b R, f R[, b], c R. Then cf R[, b] nd cf = c f Theorem [Intervl Additivity] Let, b, c R nd put them in nondecresing order: A B C. Then f R[A, C] f R[A, B] nd f R[B, C]. In this cse: c f = c f + b f Corollry. If < b nd f R[, b] then for every [c, d] [, b]: f R[c, d]. In the following theorems < b is ssumed Theorem [Monotonity] Let, b R, < b, f, g R[, b]. Suppose tht f(x) g(x) (x [, b]). Then f Theorem [ Tringle inequlity] Let, b R, < b, f R[, b]. Then f R[, b] nd f f Theorem [Men vlue Theorem] Let, b R, < b, f, g R[, b], g(x) 0 (x [, b]). Let m := inf{f(x) x b}, M := sup{f(x) x b}. g. Then m g fg M g.